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Topic: What's the product of this reaction?  (Read 11079 times)

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Offline kanonsviel

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What's the product of this reaction?
« on: September 08, 2010, 02:08:54 AM »
Is it possible to undergo a nucleophilic aromatic substitution reaction?
But, without a EDG attaching to the para or ortho position, it's unlikely to be able to form a Meisenheimer complex...
So, what kind of reactions could occur in this reaction other than a nucleophilic aromatic substitution?

Offline Jorriss

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Re: What's the product of this reaction?
« Reply #1 on: September 08, 2010, 02:11:00 AM »
It could undergo the benzyne mechanism, I forget the name, but without some intense heat I don't know if it occurs appreciably.

Offline discodermolide

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Re: What's the product of this reaction?
« Reply #2 on: September 08, 2010, 02:50:23 AM »
Is it possible to undergo a nucleophilic aromatic substitution reaction?
But, without a EDG attaching to the para or ortho position, it's unlikely to be able to form a Meisenheimer complex...
So, what kind of reactions could occur in this reaction other than a nucleophilic aromatic substitution?

You will form a deuterated benzyne which will polymerise under the reaction conditions.
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Offline kanonsviel

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Re: What's the product of this reaction?
« Reply #3 on: September 08, 2010, 05:19:28 AM »
Humm...seems like benzyne is the final answer.
but, what's the fate of these two deuteriums? does one of them remain intact if it was to form a polymer?
and the mechanism leading to a benzyne...is this an one step reaction(-OH's attack towards the deuterium, whereupon the Iodo group gets expelled??) or there is an intermediate? ???

Offline discodermolide

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Re: What's the product of this reaction?
« Reply #4 on: September 08, 2010, 05:47:03 AM »
Humm...seems like benzyne is the final answer.
but, what's the fate of these two deuteriums? does one of them remain intact if it was to form a polymer?
and the mechanism leading to a benzyne...is this an one step reaction(-OH's attack towards the deuterium, whereupon the Iodo group gets expelled??) or there is an intermediate? ???

No it seems to be concerted, de-protonation and immediate elimination of DI to give the benzyne. Benzynes can be trapped by dienophiles at very low temperature (-78°C)
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Offline kanonsviel

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Re: What's the product of this reaction?
« Reply #5 on: September 08, 2010, 10:45:16 AM »
I'm sorry to ask this persistently...but, please forgive my crappy English, what did you mean by "concerted", I have referred to a dictionary, yet still can't understand what it supposed to mean... :'(
may I take it for "Meanwhile"??
you were suggesting that this is an one-step reaction, do I get it precisely?
and about the deuterium, was my conjecture (one would remain intact) right or wrong? ???

Offline Jorriss

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Re: What's the product of this reaction?
« Reply #6 on: September 08, 2010, 10:47:15 AM »
concerted means the entire reaction occurs at once. All bonds are formed and broken simultaneously.

Offline discodermolide

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Re: What's the product of this reaction?
« Reply #7 on: September 08, 2010, 10:49:45 AM »
I'm sorry to ask this persistently...but, please forgive my crappy English, what did you mean by "concerted", I have referred to a dictionary, yet still can't understand what it supposed to mean... :'(
may I take it for "Meanwhile"??
you were suggesting that this is an one-step reaction, do I get it precisely?
and about the deuterium, was my conjecture (one would remain intact) right or wrong? ???

One deuterium should remain
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Offline kanonsviel

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Re: What's the product of this reaction?
« Reply #8 on: September 08, 2010, 10:53:31 AM »
I see... thanks for your scrupulous explanation :)

Offline discodermolide

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Re: What's the product of this reaction?
« Reply #9 on: September 08, 2010, 12:56:08 PM »
I see... thanks for your scrupulous explanation :)

This is why I am not a chemistry teacher, what explanation do you require exactly?
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Offline orgopete

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Re: What's the product of this reaction?
« Reply #10 on: September 08, 2010, 09:45:54 PM »
As written, I too expect a benzyne intermediate. It should then react with -OH, to give after protonation, a mixture of ortho and meta-deuterophenol.

Concerted means the electron movements occur all at the same time so no additional intermediate is thought to be present.
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Offline kanonsviel

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Re: What's the product of this reaction?
« Reply #11 on: September 09, 2010, 01:43:09 AM »
This is why I am not a chemistry teacher, what explanation do you require exactly?

I hope the sentence I have written is not misleading...I just translated it directly from Japanese, in order to express my gratitude. did it seem that I was requiring a further explanation? ???

As written, I too expect a benzyne intermediate. It should then react with -OH, to give after protonation, a mixture of ortho and meta-deuterophenol.

Concerted means the electron movements occur all at the same time so no additional intermediate is thought to be present.

According to my text book, it seems that this is the industrial route to produce phenol from halogenobenzene, but as discodermolide said, this reaction only occurs under a high temperature and pressure condition :)

Offline orgopete

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Re: What's the product of this reaction?
« Reply #12 on: September 09, 2010, 11:29:01 AM »
Without further discussion, I suggest it would give the mixture of phenols or no reaction. If as suggested these conditions are not harsh enough (though I have no indication of temperature or pressure), then it should be no reaction.
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Offline discodermolide

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Re: What's the product of this reaction?
« Reply #13 on: September 09, 2010, 11:36:18 AM »
Without further discussion, I suggest it would give the mixture of phenols or no reaction. If as suggested these conditions are not harsh enough (though I have no indication of temperature or pressure), then it should be no reaction.

Without further discussion!!!
Depending on the KOH concentration and temperature these are conditions that will produce benzyne, water may re-add but benzyne is so reactive that it will poylmerise rapidly. So at best you will get a mess.
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Offline orgopete

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Re: What's the product of this reaction?
« Reply #14 on: September 09, 2010, 12:21:12 PM »
Without further discussion!!!
Depending on the KOH concentration and temperature these are conditions that will produce benzyne, water may re-add but benzyne is so reactive that it will poylmerise rapidly. So at best you will get a mess.

Certainly, you could write that as your solution. If you could verify this as well, it would be even more convincing. I have neither. Being unaware of the benzyne polymerization rate and if I were attempting to perform this reaction and I was getting polymerization, I would doubtlessly slowly add the iodobenzene to the KOH to reduce the benzyne concentrations and increase the concentration of hydroxide. That should increase the bimolecular reaction rate.

This looks like a reaction scheme taken from an example of a benzyne reaction in which the objective is to predict the benzyne products. I presumed that was the point of using a dideuterated starting material.
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